We would like to apply Fourier inversion to \(w(\mathbf {x})h(r,G(\mathbf {x}))e(-\mathbf {v}\cdot \mathbf {x})\) , the integrand appearing in the definition of \(I_{q,\mathbf {a}}(\mathbf {c})\) . The function \(\mathbf {x}\mapsto h(r,G(\mathbf {x}))\)  does not have compact support. We define the smooth weight \(w_0:\mathbb {R}\rightarrow \mathbb {R}_{\geqslant 0}\)  by \(w_0(x) = \psi (x/17)\) , where \(\psi (x)\)  is the standard bump function in one variable. Then we define the smooth weight \(w_3:R^4 \rightarrow \mathbb {R}_{\geqslant 0}\)  by
\(w_3(\mathbf {x}) = {\left\lbrace \begin{array}{ll} \frac{w(\mathbf {x})}{w_0(G(\mathbf {x}))}, &\mbox{if } w_0(G(\mathbf {x})) \ne 0,\\0, & \mbox{otherwise.}\end{array}\right.}\) 
Note that \(w\)  is supported on \([-2,2]^4\) , and in this region \(G(\mathbf {x})\)  takes values in \([-16,16]\) . Therefore \(w_0(G(\mathbf {x})) \gg 1\)  for all \(\mathbf {x}\in \mathrm {supp}(w)\) . We deduce that \(\mathrm {supp}(w_3) = \mathrm {supp}(w)\)  and \(w(\mathbf {x}) = w_3(\mathbf {x})w_0(G(\mathbf {x}))\)  for all \(\mathbf {x}\in \mathbb {R}^4\) . Moreover, since all the derivatives of \(w_0(G(\mathbf {x}))\)  are \(O(1)\)  for any \(\mathbf {x}\in \operatorname{supp}(w)\) , we have \(w_3 \in S\) . Applying Fourier inversion, we obtain
\(I_{q,\mathbf {a}}(\mathbf {c}) = P_1 \cdots P_4 \int _{-\infty }^{\infty } p(\theta ) \int _{\mathbb {R}^4} w_3(\mathbf {x}) e(\theta G(\mathbf {x}) - \mathbf {v}\cdot \mathbf {x}) \,\mathrm {d}\mathbf {x}\,\mathrm {d}\theta ,\) 
where
\( p(\theta ) = \int _{-\infty }^{\infty } w_0(k)h(r,k)e(-\theta k)\,\mathrm {d}k.\) 
We have the estimate \(p(\theta ) \ll 1\) . Indeed, taking \(m=n=0\)  and \(N=2\)  in [1]} we see that
\(h(r,k) \ll r^{-1}(r^2 + \min (1,(r/|k|)^2),\) 
and so
\(p(\theta )\ll r^{-1}\left(\int _{|k|<r}1 \mathrm {d}k + \int _{r\leqslant |k|\leqslant 17}\left(\frac{r}{|k|}\right)^2\mathrm {d}k \right)\ll r^{-1}(r + r^2\cdot r^{-1})\ll 1.\) 
To deal with the inner integral in (REF ), which we denote by \(I(\theta ;\mathbf {v})\) , we divide into the cases \(5|\theta | \leqslant |\mathbf {v}|\)  and \(5|\theta |\geqslant |\mathbf {v}|\) . In the former case, we apply Lemma REF  with \(N=2\)  to obtain \(I(\theta ;\mathbf {v})\ll (\eta |\mathbf {v}|)^{-2}\) . The contribution to \(I_{q,\mathbf {a}}(\mathbf {c})\)  is
\( (\eta |\mathbf {v}|)^{-2}P_1\cdots P_4 \int _{5|\theta |\leqslant |\mathbf {v}|} 1 \mathrm {d}\theta \ll \eta ^{-2}|\mathbf {v}|^{-1}P_1\cdots P_4.\) 
In the case \(5|\theta |\geqslant |\mathbf {v}|\) , we use the arguments from [2]} to obtain
\(I(\theta ;\mathbf {v})\ll \left\lbrace \int _{\mathbb {R}^4}|\widehat{w_3}(\mathbf {y})| \mathrm {d}\mathbf {y}\right\rbrace  \sup _{\mathbf {y}\in \mathbb {R}^4}\left|\prod _{i=1}^4\int _{[-2,2]} e(\theta \epsilon _i x_i^2 +x_i(y_i \pm v_i))\mathrm {d}x_i \right|,\) 
where \(\widehat{w_3}\)  denotes the Fourier transform of \(w_3\) . The first factor on the right hand side of (REF ) is the \(L^1\) -norm of \(\widehat{w_3}\) , which we denote by \(\Vert \widehat{w_3}\Vert _{L^1}\) . The \(v_i\) 's can be absorbed into the supremum over \(\mathbf {y}\) , and so it remains to study
\(\int _{[-2,2]}e(\pm \theta x_i^2 - y_ix_i)\mathrm {d}x_i.\) 
We can write the integrand as \(e(\pm \theta \Phi (x_i))\) , where \(\Phi (x_i) = x_i^2 + x_iy_i\theta ^{-1}\) . Then \(|\Phi ^{\prime \prime }(x_i)|\geqslant 1\)  throughout the interval \([-2,2]\) , and so we can apply the second derivative test as found in [3]} to bound this integral by \(|\theta |^{-1/2}\) . Therefore
\(I(\mathbf {v};\theta ) \ll \Vert \widehat{w_3}\Vert _{L_1}|\theta |^{-2},\) 
Returning to (REF ), the contribution to \(I_{q,\mathbf {a}}(\mathbf {c})\)  from the range \(5|\theta | \geqslant |\mathbf {v}|\)  is bounded by
\(\Vert \widehat{w_3}\Vert _{L_1}P_1\cdots P_4 \int _{|\theta |\gg |\mathbf {v}|}|\theta |^{-2} \mathrm {d}\theta \ll \Vert \widehat{w_3}\Vert _{L_1}P_1\cdots P_4 |\mathbf {v}|^{-1}.\) 
Since
\(\frac{P_1\cdots P_4}{|\mathbf {v}|} = \frac{B^{3/2}q}{|A|^{1/2}}\min _{1\leqslant i\leqslant 4}\left(\frac{|a_i|^{1/2}}{|c_i|}\right),\) 
it remains only to show that \(\Vert \widehat{w}_3\Vert _{L^1} \ll \eta ^{-4}\) . We have
\(\Vert \widehat{w}_3\Vert _{L^1} = \int _{|\mathbf {x}| \leqslant \eta ^{-1}}|\widehat{w_3}(\mathbf {y})|\,\mathrm {d}\mathbf {y}+ \int _{|\mathbf {x}| \geqslant \eta ^{-1}}|\widehat{w_3}(\mathbf {y})|\,\mathrm {d}\mathbf {y}.\) 
The first integral is trivially \(O(\eta ^{-4})\) . For the second integral, we have
\( \int _{|\mathbf {x}|\geqslant \eta ^{-1}}|\widehat{w_3}(\mathbf {y})|\,\mathrm {d}\mathbf {y}\ll \int _{\begin{array}{c}|\mathbf {x}|\geqslant \eta ^{-1}\\ |y_1| = \max _i|y_i|\end{array}}|\widehat{w_3}(\mathbf {y})|\,\mathrm {d}\mathbf {y}.\) 
We can now apply integration by parts five times with respect to \(x_1\)  to obtain
\(\int _{|\mathbf {y}|\geqslant \eta ^{-1}}|\widehat{w_3}(\mathbf {y})|\,\mathrm {d}\mathbf {y}&\ll \int _{\begin{array}{c}|\mathbf {y}|\geqslant \eta ^{-1}\\|y_1|=\max _i|y_i|\end{array}}\left|\int _{\mathbb {R}^4}\frac{\partial ^5w_3(\mathbf {x})}{\partial x_1^5}\frac{e(-\mathbf {x}\cdot \mathbf {y})}{(-2\pi i y_1)^5}\mathrm {d}\mathbf {x}\right|\mathrm {d}\mathbf {y}\\&\ll \eta ^{-5}\int _{\begin{array}{c}|y_1|\geqslant \eta ^{-1}\\|y_2|,|y_3|,|y_4|\leqslant |y_1|\end{array}}y_1^{-5}\mathrm {d}\mathbf {y}\\&\ll \eta ^{-5}\int _{y_1\geqslant \eta ^{-1}}y_1^{-2}\,\mathrm {d}y_1\\&\ll \eta ^{-4}.\) 
